User:Gauge00/Pade

About Pade approximation
Given a function f and two integers m &ge; 0 and n &ge; 0, the Padé approximant of order (m, n) is the rational function


 * $$R(x)=\frac{p_0+p_1x+p_2x^2+\cdots+p_mx^m}{1+q_1 x+q_2x^2+\cdots+q_nx^n}$$

which agrees with $$f(x)$$ to the highest possible order, which amounts to
 * $$\begin{array}{rcl}

f(0)&=&R(0)\\ f'(0)&=&R'(0)\\ f(0)&=&R(0)\\ &\vdots& \\ f^{(m+n)}(0)&=&R^{(m+n)}(0)\end{array} $$.

Equivalently, if $$R(x)$$ is expanded in a Maclaurin series (Taylor series at 0), its first m + n terms would cancel the first m + n terms of $$f(x)$$, and as such
 * $$f(x)-R(x) = c_{m+n+1}x^{m+n+1}+c_{m+n+2}x^{m+n+2}+\cdots$$

The Padé approximant is unique for given m and n, that is, the coefficients $$p_0, p_1, \dots, p_m,$$ $$q_1, \dots, q_n$$ can be uniquely determined. It is for reasons of uniqueness that the zero-th order term at the denominator of $$R(x)$$ was chosen to be 1, otherwise the numerator and denominator of $$R(x)$$ would have been unique only up to multiplication by a constant.

Determination of the Pade coefficients
Here let's focus on a special Pade approximant where the linear numerator is chosen.


 * $$ \frac{1-x/k}{1 + b_1 x + b_2 x^2/2! + b_3 x^3/3!... } =

(1 + a_1 x + a_2 x^2/2! + a_3 x^3 / 3! + ... ) $$

where $$ 1, a_1, ... $$ are given variables from the function $$ 1 + a_1 x + a_2 x^2/2! + a_3 x^3 / 3! + ... $$, and $$ k $$, and $$ 1, b_1, ... $$ are the variable we have to determine.

Above equation can be rewritten as


 * $$ \left( 1 + b_1 x + \frac{1}{2!} b_2 x^2 +  \frac{1}{3!} b_3 x^3 ... \right) *

\left( 1 + a_1 x + \frac{1}{2!} a_2 x^2 + \frac{1}{3!} a_3 x^3  + ... \right) = {1-x/k} $$

Differentiating both sides and setting x=0, we get


 * $$ b_1 + a_1 = -1/k $$


 * $$ b_2 + 2 b_1 a_1 + a_2 = 0$$
 * $$ b_3 + 3 b_2 a_1 + 3 b_1 a_2 + a_3 = 0 $$
 * $$ b_4 + 4 b_3 a_1 + 6 b_2 a_2 + 4 b_1 a_3 + a_4 = 0$$
 * $$ b_n + C(n, 1) b_{n-1} a_1 + C(n, 2) b_{n-2} a_2 + ... +

C(n, n-2) b_2 a_{n-2} + C(n, n-1) b_1 a_{n-1} + C(n, n) a_n = 0 $$

(1, 0) order
In this case, $$ b_1 = b_2 = ... = 0 $$

Therefore $$ b_1 + a_1 = a_1 = -1/k $$, we can see that the numerator is the form of $$ 1 + a_1 x $$ as we expected.

Let's call this $$ k $$ as $$ k_0 $$

$$ -\frac{1}{k_0} = a_1 $$,

(1, 1) order
In this case, $$ b_2 = b_3 = 0 $$, therefore
 * $$ b_2 + 2 b_1 a_1 + a_2 = b_2 + b_1 \left( \frac{-2}{k_0} \right) + a_2

= b_1 \left( \frac{-2}{k_0} \right) + a_2 = 0 $$

Then


 * $$ (-a_1 - 1/k) \left( \frac{-2}{k_0}\right) + a_2 = 0 $$
 * $$ -a_1 - 1/k = \frac{k_0 a_2}{2}$$

Let's call this $$ k$$ as $$ k_1 $$, then
 * $$ -1/k_1 = a_1 + \frac{k_0 a_2} {2}$$

After defining $$k_1 $$ as such,


 * $$ a_2 = (a_1 + 1/k_1) \left( \frac{-2}{k_0} \right)

= -\left( \frac{2}{k_0 k_1} + \frac{2a_1}{k_0}  \right) $$


 * $$ b_2 = (- b_1 -a_1 - 1/k_1) * \left( \frac{-2}{k_0}\right) $$

(1, 2) order
In this case $$ b_3=0$$


 * $$ b_3 + 3 b_2 a_1 + 3 b_1 a_2 + a_3 = 0 $$
 * $$ b_3 + 3 a_1 (- b_1 - a_1 - 1/k) \left( \frac{-2}{k_0} \right)

+ 3 b_1 a_2 + a_3 = 0 $$
 * $$ b_3 + b_1 \left( \frac{6a_1}{k_0} + 3a_2 \right)

+ \frac{6a_1^2}{k_0} + \frac{6a_1}{k_0 k_1} + a_3 = 0; $$
 * $$ b_3 + b_1 \left( \frac{6a_1}{k_0} + 3a_2 \right)

+ 3a_1 \left( \frac{2a_1}{k_0} + \frac{2}{k_0 k_1} \right) + a_3 = 0; $$


 * $$ b_3 + b_1 \left( \frac{6}{k_0} (a_1 + a_2 k_0/2) \right)

+ \frac{3a_2}{k_0} + a_3 = 0; $$ therefore
 * $$ b_3 + b_1 \left( \frac{-6}{k_0 k_1} \right) + \frac{3a_2}{k_0} + a_3 = 0; $$

Let's define $$ k_2 $$ that makes $$ b_3 = 0 $$;


 * $$ -1/k = -1/k_2 = a_1 + \left( \frac{k_0 k_1}{6} \right)

\left( \frac{3a_2}{k_0} + a_3 \right) = a_1 + \frac{k_1a_2}{2} + \frac{k_0k_1a_3}{6}$$

After defining $$k_2 $$ as such,


 * $$ a_3 = \left( \frac{-6}{k_0k_1} \right)

\left( \frac{1}{k_2} + a_1 + \frac{k_1 a_1}{2} \right) = -\left( \frac{6}{k_0k_1k_2} + \frac{6a_1}{k_0k_1} + \frac{3a_2}{k_0} \right)$$
 * $$ b_3 = (-b_1 -a_1 - \frac{1}{k_2}) * \left( \frac{-6}{k_0 k_1} \right) $$

(1, 3) order

 * $$ b_4 + 4 b_3 a_1 + 6 b_2 a_2 + 4 b_1 a_3 + a_4 = 0$$
 * $$ b_4 + 4 a_1 (-b_1-a_1 - \frac{1}{k_2}) * \left( \frac{-6}{k_0 k_1} \right)

+ 10 a_2 (-b_1 - a_1 - \frac{1}{k_1}) * \frac{-2}{k_0} + 4 a_3 b_1 + a_4 = 0 $$
 * $$ b_4 + b_1 \left( \frac{24a_1}{k_0k_1} + \frac{12a_2}{k_0} + 4a_3 \right)

+ \frac{24a_1^2}{k_0k_1} + \frac{24a_1}{k_0k_1k_2} + \frac{12a_1 a_2}{k_0} + \frac{12a_2}{k_0k_1} + a_4 = 0$$
 * $$ b_4 + b_1 \frac{24}{k_0k_1} * \left( a_1 + \frac{a_2k_1}{2} + \frac{k_0 k_1 a_3}{6} \right)

+ 4a_1 \left( \frac{6}{k_0k_1k_2} + \frac{6a_1}{k_0k_1} + \frac{3a_2}{k_0} \right) + \frac{12a_2}{k_0k_1} + a_4 = 0 $$


 * $$ b_4 + b_1 \left( \frac{-24}{k_0k_1k_2} \right) - 4a_1a_3 + \frac{12a_2}{k_0k_1} + a_4 = 0 $$
 * $$ b_4 + b_1 \left( \frac{-24}{k_0k_1k_2} \right) + \frac{4a_3}{k_0} + \frac{12a_2}{k_0k_1} + a_4 = 0 $$

Let's define $$ k_3 $$ that makes $$ b_4 = 0 $$;
 * $$ -1/k = -1/k_3 = a_1 + (\frac{k_0k_1k_2}{24}) * \left(

\frac{4a_3}{k_0} + \frac{12a_2}{k_0k_1} + a_4 \right)           = a_1 + \frac{k_2a_2}{2} + \frac{k_1k_2a_3}{6} + \frac{k_0k_1k_2a_4}{24}$$

After defining $$k_3 $$ as such,


 * $$ a_4 = -\left( \frac{24}{k_0k_1k_2k_3} + \frac{24a_1}{k_0k_1k_2}

+ \frac{12a_2}{k_0k_1} + \frac{4a_3}{k_0} \right) $$


 * $$ b_4 = (-b_1 -a_1 - \frac{1}{k_3}) * \left( \frac{-24}{k_0 k_1k_2} \right) $$

(1, 4) order

 * $$ b_5 + 5b_4a_1 + 10b_3a_2 + 10b_2a_3 + 5b_1 a_4 + a_5 = 0 $$
 * $$ b_5 + 5a_1 (-b_1 -a_1 - \frac{1}{k_3}) * \left( \frac{-24}{k_0 k_1k_2} \right)

+ 10a_2 (-b_1 -a_1 - \frac{1}{k_2}) * \left( \frac{-6}{k_0 k_1} \right) + 10a_3 (-b_1 -a_1 - \frac{1}{k_1}) * \left( \frac{-2}{k_0} \right) + 5a_4b_1 + a_5 = 0 $$
 * $$ b_5 + b_1 \left( \frac{120a_1}{k_0k_1k_2}

+ \frac{60a_2}{k_0k_1} + \frac{20a_3}{k_0} + 5a_4 \right)      +(-5a_1) * \left( \frac{24}{k_0k_1k_2k_3} + \frac{24a_1}{k_0k_1k_2} + \frac{12a_2}{k_0k_1} + \frac{4a_3}{k_0} \right)          + \frac{60a_2}{k_0k_1k_2} + \frac{20a_3}{k_0k_1} + a_5 = 0;  $$
 * $$ b_5 + b_1 \left( \frac{-120}{k_0k_1k_2k_3} \right) + \frac{60a_2}{k_0k_1k_2} + \frac{20a_3}{k_0k_1}

-5a_1a_4 + a_5 = 0; $$
 * $$ b_5 + b_1 \left( \frac{-120}{k_0k_1k_2k_3} \right) + \frac{60a_2}{k_0k_1k_2} + \frac{20a_3}{k_0k_1}

+\frac{5a_4}{k_0} + a_5 = 0; $$

Let's define $$ k_4 $$ that makes $$ b_5 = 0 $$;
 * $$ -1/k_4 = a_1 + \left( \frac{-k_0k_1k_2k_3}{120} \right) *

\left( \frac{60a_2}{k_0k_1k_2} + \frac{20a_3}{k_0k_1} +\frac{5a_4}{k_0} + a_5 \right) = a_1 + \left( \frac{k_3a_2}{2!} + \frac{k_3a_3}{3!} + \frac{k_1k_2k_3a_4}{4!}              + \frac{k_0k_1k_2k_3a_5}{5!} \right) $$

After defining $$k_4 $$ as such,


 * $$ a_5 = -\left( \frac{5!}{k_0k_1k_2k_3k_4} + \frac{5!a_1}{k_0k_1k_2k_3}

+ \frac{5!a_2}{2!k_0k_1k_2} + \frac{5!a_3}{3!k_0k_1} + \frac{5!a_4}{4!k_0} \right) $$


 * $$ b_5 = \left( -b_1 - a_1 - \frac{1}{k_4} \right) * \left( \frac{-5!}{k_0k_1k_2k_3} \right) $$

generalizing

 * $$ b_n + C(n, 1)b_{n-1}a_1 + C(n, 2)b_{n-2}a_2 + C(n, 3)b_{n-3}a_3 + ... + C(n, n-2)b_2a_{n-2}

+ C(n, n-1)b_1a_{n-1} + a_n = 0; $$


 * $$ b_n

+ C(n, 1)\left( -b_1 -a_1 -\frac{1}{k_{n-2}} \right) \left( \frac{-(n-1)!}{k_0k_1...k_{n-3}} \right) a_1 + C(n, 2)\left( -b_1 -a_1 -\frac{1}{k_{n-3}} \right) \left( \frac{-(n-2)!}{k_0k_1...k_{n-4}} \right) a_2 $$

+ C(n, n-2) \left( -b_1 -a_1 -\frac{1}{k_{1}} \right)\left( \frac{-2!}{k_0} \right)a_{n-2} + C(n, n-1) b_1a_{n-1} + a_n = 0 $$

Then


 * $$ b_n + b_1 \left[ C(n, 1) \left( \frac{(n-1)!}{k_0k_1...k_{n-3}} \right) a_1

+ C(n, 2) \left( \frac{(n-2)!}{k_0k_1...k_{n-4}} \right) a_2 + ... + C(n, n-2) \left( \frac{2!}{k_0} \right) a_{n-2} + C(n, n-1) a_{n-1} \right] $$
 * $$ C(n, 1)\left( \frac{(n-1)!}{k_0k_1...k_{n-2}} \right) a_1

+ C(n, 1) \left( \frac{(n-1)!}{k_0k_1...k_{n-3}} \right) a_1^2 + C(n, 2) \left( \frac{(n-2)!}{k_0k_1...k_{n-4}} \right) a_1a_2 $$

+ C(n, 3) \left( \frac{(n-3)!}{k_0k_1...k_{n-5}} \right) a_1a_3 + ...   + C(n, n-2) \left( \frac{2}{k_0} \right) a_1 a_{n-2}  $$

C(n, 2) \left( \frac{(n-2)!}{k_0k_1...k_{n-3}} \right) a_2 + C(n, 3) \left( \frac{(n-3)!}{k_0k_1...k_{n-4}} \right) a_3 + ...  + C(n, n-2) \left( \frac{2}{k_0k_1} \right) a_{n-2} + a_n = 0 $$

Since


 * $$ C(n,1)(n-1)!= n!; C(n,1)(n-2)!= \frac{n!}{2!}; $$

Then


 * $$ C(n, 1) \left( \frac{(n-1)!}{k_0k_1...k_{n-3}} \right) a_1

+ C(n, 2) \left( \frac{(n-2)!}{k_0k_1...k_{n-4}} \right) a_2 + ... + C(n, n-2) \left( \frac{2!}{k_0} \right) a_{n-2} + C(n, n-1) a_{n-1} $$
 * $$ = \frac{n!}{k_0k_1..k_{n-3}}a_1 + \frac{n!}{2! k_0k_1...k_{n-4}}a_2 + ...

+ \frac{n! }{(n-2)!k_0} a_{n-2} + \frac{n! }{(n-1)!} a_{n-1} $$
 * $$ = \left( \frac{n!}{k_0k_1..k_{n-3}} \right) *

\left( a_1 + \frac{k_{n-3}a_2}{2!} + \frac{k_{n-3}k_{n-2}a_3}{3!} +            ... + \frac{k_1k_2...k_{n-3}a_{n-2}}{(n-2)!}                  + \frac{k_0k_1k_2...k_{n-3}a_{n-1}}{(n-1)!}    \right) $$
 * $$ = \left( \frac{n!}{k_0k_1..k_{n-3}} \right) \left( \frac{-1}{k_{n-2}} \right)

= \frac{-n!}{k_0k_1..k_{n-2}} $$

And


 * $$ C(n, 1)\left( \frac{(n-1)!}{k_0k_1...k_{n-2}} \right) a_1

+ C(n, 1) \left( \frac{(n-1)!}{k_0k_1...k_{n-3}} \right) a_1^2 + C(n, 2) \left( \frac{(n-2)!}{k_0k_1...k_{n-4}} \right) a_1a_2 $$

+ C(n, 3) \left( \frac{(n-3)!}{k_0k_1...k_{n-5}} \right) a_1a_3 + ...   + C(n, n-2) \left( \frac{2}{k_0} \right) a_1 a_{n-2}  $$
 * $$ = \frac{n!}{k_0k_1..k_{n-2}} a_1 + \frac{n!}{k_0k_1..k_{n-3}} a_1^2

+ \frac{n!}{2! k_0k_1...k_{n-4}} a_1a_2 + ...   + \frac{n! }{(n-2)!k_0} a_1a_{n-2} $$
 * $$ = na_1 \left( \frac{(n-1)!}{k_0k_1...k_{n-2}} + \frac{(n-1)!}{k_0k_1...k_{n-3}} a_1

+ \frac{(n-1)!}{2! k_0k_1...k_{n-4}} a_2 + ... + \frac{(n-1)! }{(n-2)!k_0} a_{n-2} \right) $$
 * $$ = na_1 (-a_{n-1}) = \frac{na_{n-1}}{k_0} $$

Therefore
 * $$ b_n + \left( \frac{-n!}{k_0k_1..k_{n-2}} \right) b_1 + \frac{na_{n-1}}{k_0}

+    C(n, 2) \left( \frac{(n-2)!}{k_0k_1...k_{n-3}} \right) a_2 + C(n, 3) \left( \frac{(n-3)!}{k_0k_1...k_{n-4}} \right) a_3 + ...  + C(n, n-2) \left( \frac{2}{k_0k_1} \right) a_{n-2} + a_n = 0 $$


 * $$ b_n + \left( \frac{-n!}{k_0k_1..k_{n-2}} \right) b_1

+ \left(      \frac{n!}{2!k_0k_1...k_{n-3}} a_2      + \frac{n!}{3!k_0k_1...k_{n-4}} a_3 + ...     + \frac{n!}{(n-2)!k_0k_1} a_{n-2}      + \frac{n!}{(n-1)!k_0} a_{n-1} + a_n \right) = 0; $$

Solving the last equation, for the nth order Pade approximant $$ b_n = 0 $$


 * $$ -1/k_{n-1} = a_1 + \left( \frac{k_0k_1..k_{n-2}}{n!} \right)

\left( \frac{n!}{2!k_0k_1...k_{n-3}} a_2     + \frac{n!}{3!k_0k_1...k_{n-4}} a_3 + ...     + \frac{n!}{(n-2)!k_0k_1} a_{n-2}      + \frac{n!}{(n-1)!k_0} a_{n-1} + a_n   \right) $$
 * $$ = a_1 + \frac{k_{n-2}} {2!} a_2 + \frac{k_{n-2}k_{n-3}} {3!} a_3

+ ...        + \frac{k_{n-2}k_{n-3}...k_2} {(n-2)!} a_{n-2} + \frac{k_{n-2}k_{n-3}...k_2k_1} {(n-1)!} a_{n-1} + \frac{k_{n-2}k_{n-3}...k_2k_1k_0} {n!} a_n

$$

A simple application
For example, for a given function $$ f(x) = 1 - 3x + 5x^2 - 7x^3 = 1-3x + 10x^2/2! - 42x^3/3! + $$.


 * $$ 0 = 1 - 3x; $$ and $$ x=1/3=k_0 $$

Therefore the (1, 0) order Pade appromant is $$ (1-x/k_0) $$, that is $$ 1 - 3x $$.

And


 * $$ 0 = 1 - 3x + 5x(1/3); $$ and $$ 0 = 1 - 4x/3; $$ and $$x=3/4=k_1 $$
 * $$ b_1 = -a_1 -1/k_1 = 3 -1/(3/4) = 3 -4/3 = 5/3 $$

Therefore the (1, 1) order Pade approximant is the form of


 * $$ \frac{1-x/k_1}{1+b_1x} = \frac{1-4x/3}{1+5x/3} $$

And
 * $$ 0 = 1 - 3x + 5x(3/4) - 7x(3/4)(1/3) = 1 - 3x +2x= 1 - x = 0; $$ and $$x = 1 = k_2 $$
 * $$ b_1 = -a_1 - 1/k_2 = 3 - 1 = 2$$
 * $$ b2 = (-b1 - a1- 1/k_1)*(-2/k_0) = (-2 + 3 -4/3)*(-2/(1/3)) = (-1/3)*(-6) = 2$$

Therefore the (1, 2) order Pade approximant is the form of


 * $$ \frac{1-x/k_2}{1+b_1x+b_2x^2} = \frac{1-x}{1+2x+2x^2}$$

When $$a_0$$ is not one

Let's calculate $$ k_0, k_1, k_2, k3$$ in the case of $$ 2f(x) = 2*(1 - 3x + 5x^2 - 7x^3) = 2 - 6x + 10x^2 - 14x^3$$.

$$ 0 = 2 - 6x $$ Therefore $$ k_0 = 1/3 $$.

$$ 0 = 2 - 6x + 10x(1/3) = 2 - 8x/3$$ Therefore $$ k_1 = 3/4 $$

$$ 0 = 2 - 6x + 10x(3/4) - 14x(3/4)(1/3) = 2 - 6x + 15x/2 - 7x/2 = 2 -2x $$ Therefore $$ k_2 = 1 $$

As a result, we can see that $$ k_0, k_1, ... $$ can be computed using the same method.

Householder's method
Let's see some properties of the derivatives of $$ 1/f(x) $$ for a function $$ f(x) $$

Derivatives of the $$D=1/f$$
Let's define a new function $$D=1/f(x)$$. Then $$1 = f(x)D(x) $$

By defferentiating


 * $$ 0 = f^\prime (x) D(x) + f(x)D^\prime (x) $$, that is,


 * $$-fD^{\prime} = Df^{\prime} $$

Differentiating multiple times, we get


 * $$-fD^{\prime\prime} = 2D^{\prime} f^\prime + Df^{\prime \prime} $$


 * $$-fD^{\prime\prime\prime} = 3D^{\prime\prime}f^{\prime} + 3D^{\prime}f^{\prime\prime}

+ Df^{\prime\prime\prime}$$


 * $$-fD^{\prime\prime\prime\prime} = 4D^{\prime\prime\prime}f^{\prime} + 6D^{\prime\prime} f^{\prime\prime}

+ 4D^{\prime}f^{\prime\prime\prime} + Df^{\prime\prime\prime\prime} $$

The coefficients of the above equations are those of the Pascal's triangle. Taking $$C(n, k)$$ notation for the binomial coefficient, we would get
 * $$-fD^{(n)} = C(n, 1)D^{(n-1)}f^{\prime} + C(n, 2) D^{(n-2)} f^{\prime\prime}

+ C(n, 3) D^{(n-3)} f^{\prime\prime\prime} + ... + C(n, n) Df^{(n)} $$

A few examples of $$ D(x) = 1/f(x) $$ derivatives
Though some of followings would not be used at the derivation, let's expand some of above equations to see what forms they have,


 * $$ -fD = -1 $$


 * $$ -fD^{\prime} = f^{\prime}/f $$


 * $$\begin{array}{rl}

-fD^{\prime\prime} = &2D^{\prime}f^{\prime} + f^{\prime\prime}D \\ = & 2 \left( -f^{\prime} /{f^2} \right) f^{\prime} + f^{\prime\prime}/f \\ = &\left( 1/f^2 \right) \left( -2{f^{\prime}} ^2 + ff^{\prime\prime} \right) \end{array} $$


 * $$ -fD^{\prime\prime\prime} = ( 1/f^3 )

(  6{f^\prime}^3 -6ff^\prime f^{\prime\prime} + f^2 f^{\prime\prime\prime} ) $$


 * $$ -fD^{\prime\prime\prime\prime} = ( 1/f^4 )

( -24{f^\prime}^4 + 36f{f^\prime}^2 f^{\prime\prime} - 8 f^2 f^\prime f^{\prime\prime\prime} -6f^2 {f^{\prime\prime}}^2 + f^3 f^{\prime\prime\prime\prime} ) $$

...

Therefore


 * $$ D = (-1/f) (-1) $$


 * $$ D^{\prime} = (-1/f^2) f^{\prime} $$


 * $$ D^{\prime\prime} = ( -1/f^3 ) ( -2{f^{\prime}} ^2 + ff^{\prime\prime} )$$


 * $$ D^{\prime\prime\prime} = ( -1/f^4 )

(  6{f^\prime}^3 -6ff^\prime f^{\prime\prime} + f^2 f^{\prime\prime\prime} ) $$


 * $$ D^{\prime\prime\prime\prime} = ( -1/f^5 )

( -24{f^\prime}^4 + 36f{f^\prime}^2 f^{\prime\prime} - 8 f^2 f^\prime f^{\prime\prime\prime} -6f^2 {f^{\prime\prime}}^2 + f^3 f^{\prime\prime\prime\prime} ) $$

A Derivation of the Householder's method
By the Taylor expansion,


 * $$ f(x) = f(x_0)

+ f^\prime (x_0) (x-x_0) + (1/2!)f^{\prime\prime} (x_0) {(x-x_0) }^2 + (1/3!)f^{\prime\prime\prime} (x_0) {(x-x_0)}^3 + ... $$

where, $$x_0$$ is the initial guess of the root of $$ f(x) = 0$$.

Newton's method

Let's call the (1, 0) order Pade approximant of $$ f(x) $$ as $$ f_1(x) $$, then $$ f_1(x)$$ has root at the point $$ k_0 $$, where


 * $$ k_0 = -a_0/a_1 = -f(x_0)/f^\prime (x_0) $$, that is,


 * $$ x_1 - x_0 = - f(x_0)/f^{\prime} (x_0) $$

This is the Newton's method. This is identical to the solution of $$ 0 = f(x_0) + f^\prime (x_0) (x_1 - x_0)$$

Let's call above $$ x_1 - x_0 $$ as $${\Delta}_{Newton}$$ or $$\Delta_{(1)}$$


 * $$\begin{array}{rl}

{\Delta}_{Newton} = \Delta_{(1)} = & - f(x_0)/f^{\prime} (x_0) = -f/f^{\prime} = D(x_0)/D^{\prime}(x_0) \end{array} $$

Halley's method

Let's call the (1, 1) order Pade approximant of $$ f(x) $$ as $$ f_2(x) $$, then $$ f_2(x)$$ has root at the point $$ k_1 = x_1 - x_0$$, where


 * $$ -1/k_1 = \frac{a_1}{a_0} + \frac{k_0 a_2} {2a_0}$$, that is


 * $$ x_1 - x_0 = k_1 = -\frac{a_0} {a_1 + k_0 a_2/2} $$
 * $$ = \frac{f(x_0)} {f^\prime (x_0)  + {\Delta}_{Newton} f^{\prime\prime}  (x_0) /2 }

= \frac{f(x_0)} {f^\prime (x_0) - f (x_0) f^{\prime\prime}  (x_0) /2f^\prime (x_0) }  $$


 * $$ = \frac{-2f(x_0) f^\prime (x_0) } {2{f^\prime}^2 (x_0)  - f (x_0) f^{\prime\prime} (x_0)  }  $$

This is the Halley's method. This is identical to the solution of
 * $$ 0 = f(x_0) + f^{\prime} (x_0) (x-x_0) + (1/2!) f^{\prime\prime} (x_0) (x-x_0) {\Delta}_{newton} $$

And Let's call $$x_1 - x_0$$ as $$ {\Delta}_{Halley}$$ or $$\Delta_{(2)}$$



{\Delta}_{Halley} = \Delta_{(2)} = 2ff^{\prime} / ( f^{\prime\prime}f - 2{f^\prime}^2 ) = 2 (f^\prime/f)/ \left[ (1/f^2) ( f^{\prime\prime}f - 2{f^\prime}^2 ) \right] = 2 (-fD^{\prime} )/(-fD^{\prime\prime}) = 2 D^{\prime}/D ^{\prime\prime} $$

We shall see that the Halley's method is the second kind of Householder's method.

(1,2) order Pade approximation

We can compute $$ k_2$$, the root of the (1,2) order Pade Approximant, by solving


 * $$ -1/k_2 = a_1/a_0 + \frac{k_1a_2/a_0}{2} + \frac{k_0k_1a_3/a_0}{6}$$

Which is identical to solving the following equation,


 * $$ 0 = f(x_0)

+ f^\prime (x_0) (x-x_0) + (1/2!)f^{\prime\prime} (x_0) (x-x_0) k_1 + (1/3!)f^{\prime\prime\prime} (x_0) (x-x_0) k_1 k_0$$, that is,


 * $$ 0 = f(x_0)

+ f^\prime (x_0) (x-x_0) + (1/2!)f^{\prime\prime} (x_0) (x-x_0) {\Delta}_{Halley} + (1/3!)f^{\prime\prime\prime} (x_0) (x-x_0) {\Delta}_{Halley} {\Delta}_{Newton} $$

Therefore,


 * $$ \begin{array}{rl}

\Delta_{(3)} = & ( -f ) / \left[ f^{\prime} + (1/2!) f^{\prime\prime} {\Delta}_{Halley} + (1/3!)f^{\prime\prime\prime} {\Delta}_{Halley} {\Delta}_{Newton} \right] \\[0.7em] = & (-f)/ \left[ f^{\prime} + (1/2!) f^{\prime\prime} (2 D^{\prime}/D ^{\prime\prime} ) + (1/3!)f^{\prime\prime\prime} (2 D^{\prime}/D ^{\prime\prime} ) (D/D^{\prime} ) \right] \\[0.7em]

= & (-3f)/ \left[ 3f^{\prime} + 3 f^{\prime\prime} (D^{\prime}/D ^{\prime\prime} ) + f^{\prime\prime\prime} (D/D ^{\prime\prime} ) \right] \\[0.7em] = & (-3f D ^{\prime\prime} ) / \left[ 3f^{\prime} D ^{\prime\prime} + 3 f^{\prime\prime} D^{\prime} + f^{\prime\prime\prime} D \right] \\[0.7em]

= & (-3f D ^{\prime\prime} ) / (-f D ^{\prime\prime\prime} ) \\[0.7em] = & 3D ^{\prime\prime} / D ^{\prime\prime\prime} \\[0.7em]

\end{array} $$ This is the third kind of Householder's method.

Now generally,

(1, n-1) Pade approximation

We can compute $$ k_{n-1}$$, the root of the (1,n-1) order Pade Approximant, by solving


 * $$ -1/k_{n-1} = a_1/a_0 + \frac{k_{n-2}} {2!} a_2/a_0

+ \frac{k_{n-2}k_{n-3}} {3!} a_3/a_0 + ...        + \frac{k_{n-2}k_{n-3}...k_2} {(n-2)!} a_{n-2}/a_0 + \frac{k_{n-2}k_{n-3}...k_2k_1} {(n-1)!} a_{n-1}/a_0 + \frac{k_{n-2}k_{n-3}...k_2k_1k_0} {n!} a_n/a_0

$$

Which is identical to solving the following equation,


 * $$ 0 = f(x_0)

+ f^\prime (x_0) (x-x_0) + (1/2!)f^{\prime\prime} (x_0) (x-x_0) k_{n-2} + (1/3!)f^{\prime\prime\prime} (x_0) (x-x_0) k_{n-2}k_{n-3} + (1/4!)f^{\prime\prime\prime} (x_0) (x-x_0) k_{n-2}k_{n-3} k_{n-4} + ... $$,

that is,


 * $$ 0 = f(x_0)

+ f^\prime (x_0) (x-x_0) + (1/2!)f^{\prime\prime} (x_0) (x-x_0) {\Delta}_{n-1} + (1/3!)f^{\prime\prime\prime} (x_0) (x-x_0) {\Delta}_{n-1} {\Delta}_{n-2} + (1/4!)f^{\prime\prime\prime} (x_0) (x-x_0) {\Delta}_{n-1} {\Delta}_{n-2} {\Delta}_{n-3} + ...$$

Therefore,


 * $$ \begin{array}{rl} \Delta_{(n)} = & (-f) / \left[ f^{\prime} + (1/2!) f^{\prime\prime} {\Delta}_{(n-1)}

+ (1/3!)f^{\prime\prime\prime} {\Delta}_{(n-1)} {\Delta}_{(n-2)} + ... \right]\\[0.7em]

= & (-f)/ \left[ f^{\prime} + (1/2!) f^{\prime\prime} ((n-1) D^{(n-2)}/D^{ (n-1)} ) + (1/3!)f^{\prime\prime\prime} ((n-1) D^{(n-2)}/D^{ (n-1)} ) ((n-2) D^{(n-3)}/D^{ (n-2)} ) + (1/4!)f^{\prime\prime\prime\prime} ((n-1) D^{(n-2)}/D^{ (n-1)} ) ((n-2) D^{(n-3)}/D^{ (n-2)} ) ((n-3) D^{(n-4)}/D^{ (n-3)} ) + ...\right] \\[0.7em]

= & (-f)/ \left[ f^{\prime} + (1/2!) f^{\prime\prime} ((n-1) D^{(n-2)}/D^{ (n-1)} ) + (1/3!)f^{\prime\prime\prime} ((n-1)(n-2) D^{(n-3)}/D^{ (n-1)} ) + (1/4!)f^{\prime\prime\prime\prime} ((n-1)(n-2)(n-3) D^{(n-4)}/D^{ (n-1)} ) + ... + ((n-1)!/n!) f^{(n)}(D/D^{(n-1)}) \right] \\[0.7em]

= & (-nf D^{ (n-1)} )/ \left[ C(n, 1) f^{\prime} D^{ (n-1)} + C(n, 2) f^{\prime\prime}  D^{(n-2)} + C(n, 3) f^{\prime\prime\prime} D^{(n-3)} + C(n, 4) f^{\prime\prime\prime\prime} D^{(n-4)} + ... + C(n, n) f^{(n)} D \right] \\[0.7em] = & (-nf D^{ (n-1)} / (-f D^{(n)}) \\[0.7em] = & n D^{ (n-1)} /  D^{(n)}  \\[0.7em]

\end{array} $$

This is the Householder's method